We can use Newton's Law of Cooling: The temperature T of an object with intial temperature `T_0` after t minutes in a room with ambient temperature `T_r` is given by `T=Ce^(-kt)+T_r` .
Here `T_r=0` . We are given points `(t,T)` as `(0,100);(20,50)`
`100=Ce^(0t)==>C=100`
`50=100e^(-20k)`
`e^(-20k)=.5`
`-20k=ln(.5)`
`k=(ln(.5))/-20~~.035`
So `T=100e^(-.035t)` gives the temperature at time t:
(a) If the temperature of the bar is `25^@` find the time t:
`25=100e^(-.035t)`
`e^(-.035t)=.25`
`t=(ln(.25))/(-.035)~~39.6`
So the time required is approximately 39.6 minutes
(b) Find the temperature after 10 minutes:
`T=100e^(-.035*10)~~70.5`
So the temperature at 10 minutes is approximately `70.5^@`